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Pennies, Dimes, and Quarters (Posted on 2009-02-17) Difficulty: 2 of 5
In my left pocket I have a mix of pennies (1 cent) and quarters (25 cents). In my right pocket I have a bunch of dimes (10 cents). The number of coins in each pocket is the same, so is the cash value. What is the smallest (nonzero) number of coins I can have?

What if I had coins of x cents and z cents in my left pocket and coins of y cents in my right pocket - is there some quantitiy of coins I can have so that each pocket has the same number of coins and same cash value? (The value of the coins are positive integers x > y > z.)

See The Solution Submitted by Brian Smith    
Rating: 2.0000 (1 votes)

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Another approach | Comment 9 of 10 |
Start with two piles. Each time you put a dime in the right pile, put a penny or quarter in the left pile.

Adding a penny and a dime will make the right pile have 9 more cents. Adding a quarter and a dime will make the left pile have 15 more cents. Since you want the difference to be 0, you solve 9P=15Q, or P=5, Q=3, thus D=8. (9P=15Q is solved by noting 9P=15Q=LCM(9,15) is the least possible solution, thus Q=LCM(9,15)/15 and P=LCM(9,15)/9)

One can use a similar approach for part 2. If you have A coins of x cents, B coins of y cents, and C coins of z cents, then (y-z)C=(x-y)A=LCM(y-z, x-y) thus C=LCM(y-z,x-y)/(y-z) and A=LCM(y-z,x-y)/(x-y)

  Posted by Gamer on 2009-02-18 04:17:59
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